Question

For the ideal solution formed by two liquids A and B which of the
following is correct?
(Here XA & Xg represent mole fraction of solvent & solute in
liquid phase respectively and YA & Yp represents mole fraction of
solvent & solute in vapour phase respectively. Ptotal is total
vapour pressure of solution).​

Answer 1

Let $\small\tt {P^0_A}$ and $\small\tt {P^0_B}$ be vapour pressure of pure liquids A and B respectively.

Then total vapour pressure of the solution is given by,

$\small\text{ \longrightarrow\tt{P_{Total}=P^0_A\,X_A+P^0_B\,X_B\quad\quad\dots(1)} }$

Taking $\small\tt{X_B=1-X_A}$ in (1),

$\small\text{ \longrightarrow\tt{P_{Total}=P^0_A\,X_A+P^0_B\,(1-X_A)} }$

$\small\text{ \longrightarrow\tt{P_{Total}=P^0_A\,X_A+P^0_B-P^0_B\,X_A} }$

$\small\text{ \longrightarrow\tt{P_{Total}=(P^0_A-P^0_B)\,X_A+P^0_B\quad\quad\dots(2)} }$

This equation is in the form $\small\tt {y=mx+c}$ where $\small\text{ \tt {y=P_{Total}} }$ and $\small\tt {x=X_A.}$

Hence plot of $\small\text{ \tt {P_{Total}} }$ versus $\small\tt {X_A}$ is linear.

Taking $\small\tt{X_A=1-X_B}$ in (1),

$\small\text{ \longrightarrow\tt{P_{Total}=P^0_A\,(1-X_B)+P^0_B\,X_B} }$

$\small\text{ \longrightarrow\tt{P_{Total}=P^0_A-P^0_A\,X_B+P^0_BX_B} }$

$\small\text{ \longrightarrow\tt{P_{Total}=(P^0_B-P^0_A)\,X_B+P^0_A} }$

This equation is in the form $\small\tt {y=mx+c}$ where $\small\text{ \tt {y=P_{Total}} }$ and $\small\tt {x=X_B.}$

Hence plot of $\small\text{ \tt {P_{Total}} }$ versus $\small\tt {X_B}$ is linear.

Mole fraction of A in vapour phase is given by,

$\small\text{ \longrightarrow\tt{Y_A=\dfrac{P^0_A\,X_A}{P_{Total}}\quad\quad\dots(3)} }$

But from (2),

$\small\text{ \longrightarrow\tt{X_A=\dfrac{P_{Total}-P^0_B}{P^0_A-P^0_B}} }$

Then (3) becomes,

$\small\text{ \longrightarrow\tt{Y_A=\dfrac{P^0_A}{P_{Total}}\left(\dfrac{P_{Total}-P^0_B}{P^0_A-P^0_B}\right)} }$

$\small\text{ \longrightarrow\tt{Y_A=\dfrac{P^0_A}{P^0_A-P^0_B}\left(1-\dfrac{P^0_B}{P_{Total}}\right)} }$

$\small\text{ \longrightarrow\tt{Y_A=-\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}+\dfrac{P^0_A}{P^0_A-P^0_B}\quad\quad\dots(4)} }$

$\small\text{ \longrightarrow\tt{\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}=-Y_A+\dfrac{P^0_A}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}=\dfrac{P^0_A-Y_A(P^0_A-P^0_B)}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{P_{Total}=\dfrac{P^0_A\,P^0_B}{(P^0_B-P^0_A)Y_A+P^0_A}} }$

This equation is in the form $\small\text{ \tt {y=\dfrac{m}{ax+b}} }$ where $\small\text{ \tt {y=P_{Total}} }$ and $\small\tt {x=Y_A.}$

Hence plot of $\small\text{ \tt {P_{Total}} }$ versus $\small\tt {Y_A}$ is non - linear.

From (4), since $\small\displaystyle\tt{Y_A=1-Y_B,}$

$\small\text{ \longrightarrow\tt{1-Y_B=-\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}+\dfrac{P^0_A}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{Y_B=\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}+1-\dfrac{P^0_A}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{Y_B=\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}-\dfrac{P^0_B}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}=Y_B+\dfrac{P^0_B}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{\dfrac{P^0_A\,P^0_B}{P_{Total}(P^0_A-P^0_B)}=\dfrac{(P^0_A-P^0_B)Y_B+P^0_B}{P^0_A-P^0_B}} }$

$\small\text{ \longrightarrow\tt{P_{Total}=\dfrac{P^0_AP^0_B}{(P^0_A-P^0_B)Y_B+P^0_B}} }$

This equation is in the form $\small\text{ \tt {y=\dfrac{m}{ax+b}} }$ where $\small\text{ \tt {y=P_{Total}} }$ and $\small\tt {x=Y_B.}$

Hence plot of $\small\text{ \tt {P_{Total}} }$ versus $\small\tt {Y_B}$ is non - linear.

Hence 3rd option is correct.

Answer 2

Answer:

it is plot of Ptotal versus Xb is non linear

so, correct option is 4 th option